Summable series and the Riemann rearrangement
theorem
Jordan Bell
[email protected]
Department of Mathematics, University of Toronto
May 4, 2015
1
Introduction
Let N be the set of positive integers. A function from N to a set is called a
sequence. If X is a topological space and x ∈ X, a sequence a : N → X is said
to converge to x if for every open neighborhood U of x there is some NU such
that n ≥ NU implies that an ∈ U . If there is no x ∈ X for which a converges
to x, we say that a diverges.
Pn
Let a : N → R. We define s(a) : N → R by sn (a) = k=1 ak . We call sn (a)
the nth partial sum of the sequence a, and we call the sequence s(a) a series. If
there is some σ ∈ R such that s(a) converges to σ, we write
∞
X
ak = σ.
k=1
2
Goldbach
Euler [22, §110]: “If, as is commonly the case, we take the sum of a series to be
the aggregate of all of its terms, actually taken together, then there is no doubt
that only infinite series that converge continually closer to some value, the more
terms we actually add, can have sums”.
Euler Goldbach correspondence nos. 55, 161, 162.
3
Dirichlet
In 1837 Dirichlet proved that one can rearrange terms in an absolutely convergent series and not change the sum, and gave examples to show that this was
not the case for conditionally convergent series.
If a is a sequence and the series s(|a|) converges, we say that the series s(a)
is absolutely convergent. Because R is a complete metric space, a series being
absolutely convergent implies that it is convergent.
1
Dirichlet [14] and [15, p. 176, §101]
Elstrodt [20] and [19]
In the following theorem we prove that if a series converges absolutely, then
every rearrangement of it converges to the same value. Our proof follows Landau
[45, p. 157, Theorem 216].
Theorem 1. If a is a sequence for which s(a) converges absolutely and
∞
X
an = σ,
n=1
then for any bijection λ : N → N, the series s(a ◦ λ) converges to σ.
Proof. Let > 0, and let M be large enough so that
∞
X
|an | < .
n=M
Let r be large enough so that
{n : 1 ≤ n < M } ⊆ {λn : 1 ≤ n ≤ r}.
Fix m ≥ r, and let h : N → N be the sequence whose terms are the elements of
N \ {λn : 1 ≤ n ≤ m}
arranged in ascending order. If t + m ≥ max1≤n≤m λn then
{λn : 1 ≤ n ≤ m} ∪ {hn : 1 ≤ n ≤ t} = {n : 1 ≤ n ≤ t + m},
and hence
m
X
aλn +
n=1
t
X
ahn =
n=1
Taking t → ∞, we get
m
X
aλn +
n=1
t+m
X
an .
n=1
∞
X
ahn = σ;
n=1
the series s(a ◦ h) converges because for sufficiently large n, hn = n. Hence, for
every m ≥ r,
m
∞
∞
∞
X
X
X
X
aλn − σ = ahn ≤
|ahn | ≤
|an | < δ,
n=1
n=1
n=1
which shows that s(a ◦ λ) converges to σ.
2
n=M
4
Riemann rearrangement theorem
If a : N → R and λ : N → N is a bijection, we call the sequence a ◦ λ : N → R a
rearrangement of the sequence a.
Because N is a well-ordered set, if there are at least n elements in the set
{k ∈ N : ak ≥ 0} then it makes sense to talk about the nth nonnegative term in
the sequence a. If a were not a function from N to R but merely a function from
a countable set to R, it would not make sense to talk about the nth nonnegative
term in a or the nth negative term in a.
Riemann [60, pp. 96-97]
Our proof follows Landau [45, p. 158, Theorem 217].
Theorem 2 (Riemann rearrangement theorem). If a : N → R and s(a) converges but s(|a|) diverges, then for any nonnegative real number σ there is some
rearrangement b of a such that s(b) → σ.
Proof. Define p, q : N → R by
pn =
|an | + an
,
2
qn =
|an | − an
.
2
pn and qn are nonnegative, and satisfy pn − qn = an , pn + qn = |an |. If one of
s(p) or s(q) converges and the other diverges, we obtain a contradiction from
sn (a) =
n
X
k=1
ak =
n
X
(pk − qk ) =
k=1
n
X
pk −
k=1
n
X
qk = sn (p) − sn (q)
k=1
and the fact that s(a) converges. If both s(p) and s(q) converge, then we obtain
a contradiction from
sn (|a|) =
n
X
k=1
|ak | =
n
X
(pk + qk ) =
n
X
pk +
k=1
k=1
n
X
qk = sn (p) + sn (q)
k=1
and the fact that s(|a|) diverges. Therefore, both s(p) and s(q) diverge.
Because s(a) converges and s(|a|) diverges, there are infinitely many n with
an > 0 and there are infinitely many n with an < 0. Let Pn be the nth
nonnegative term in the sequence a, and let Qn be the absolute value of the nth
negative term in the sequence a. The fact that s(p) diverges implies that s(P )
diverges, and the fact that s(q) diverges implies that s(Q) diverges.
Let σ ≥ 0. We define sequences µ, ν : N → N by induction as follows. Let
µ1 be the least element of N such that
sµ1 (P ) > σ,
and with µ1 chosen, let ν1 be the least element of N such that
sµ1 (P ) − sν1 (Q) < σ.
3
Let m2 be the least element of N such that
sµ2 (P ) − sν1 (Q) > σ,
and with µ2 chosen, let ν2 be the least element of N such that
sµ2 (P ) − sν2 (Q) < σ.
It is straightforward to check that µ2 > µ1 and ν2 > ν1 .
Suppose that µ1 , . . . , µn and ν1 , . . . , νn have been chosen, that µn is the least
element of N such that
sµn (P ) − sνn−1 (Q) > σ,
that νn it the least element of N such that
sµn (P ) − sνn (Q) < σ
and that µn > µn−1 and νn > νn−1 . Let µn+1 be the least element of N such
that
sµn+1 (P ) − sνn (Q) > σ,
and with µn+1 chosen, let νn+1 be the least element of N such that
sµn+1 (P ) − sνn+1 (Q) < σ.
It is straightforward to check that µn+1 > µn and νn+1 > νn .
Define b : N → R by taking bn to be the nth term in
P1 , . . . , Pµ1 , −Q1 , . . . , −Qν1 , Pµ1 +1 , . . . , Pµ2 , −Qν1 +1 , . . . , −Qν2 , . . . ,
which, because the sequences µ and ν are strictly increasing, is a rearrangement
of the sequence a.
5
Symmetry
Don’t use order where it is accidental.
6
Nets
A directed set is a set D and a binary relation satisfying
• if m, n, p ∈ D, m n, and n p, then m p
• if m ∈ D, then m m
• if m, n ∈ D, then there is some p ∈ D such that m p and n p.
4
For example, let A be a set, let D be the set of all subsets of A, and say that
F G when F ⊆ G. Check that (D, ) is a directed set: for F, G ∈ D, we have
F ∪ G ∈ D, and F ∪ G is an upper bound for both F and G.
A net is a function from a directed set (D, ) to a set X. Let (X, τ ) be a
topological space, let S : (D, ) → (X, τ ) be a net, and let x ∈ X. We say that
S converges to x if for every U ∈ τ with x ∈ U there is some NU ∈ D such that
NU i implies that S(i) ∈ U . One proves that a topological space is Hausdorff
if and only if every net in this space converges to at most one point [41, p. 67,
Theorem 3].
A net S : (D, ) → R is said to be increasing if m n implies that S(m) ≤
S(n).
Lemma 3. If S : (D, ) → R is an increasing net and the range R of S has an
upper bound, then S converges to the supremum of R.
Proof. Because R is a subset of R that has an upper bound, it has a supremum,
call it σ. To say that σ is the supremum of R means that for all r ∈ R we have
r ≤ σ (σ is an upper bound) and that for all > 0 there is some r ∈ R with
σ − < r (nothing less than σ is an upper bound). Take > 0. There is some
r ∈ R with σ − < r . As r ∈ R, there is some n ∈ D with S(n ) = r . If
n n, then because S is increasing, S(n ) ≤ S(n), and hence
σ − < r = S(n ) ≤ S(n).
But S(n) ∈ R, so S(n) ≤ σ. Hence n n implies that |S(n) − σ| < , showing
that S converges to σ.
7
Unordered sums
Let A be a set, and let P0 (A) be the set of all finite subsets of A. Check that
(P0 (A), ⊆) is a directed set: if F, G ∈ P0 (A) then F ∪ G ∈ P0 (A) and F ∪ G
is an upper bound for both F and G. Let f : A → R be a function, and define
Sf : P0 (A) → R by
X
Sf (F ) =
f (a),
F ∈ P0 (A).
a∈F
If the net Sf converges, we say that the function f is summable, and we call the
element of R to which Sf converges the unordered sum of f , denoted by
X
f (a).
a∈A
If B is a subset of A, we say that f is summable over B if the restriction of f
to B is summable. If fB is the restriction of f to B and f is summable over B
(i.e. fB is summable), by
X
f (a)
a∈B
5
we mean
X
fB (a).
a∈B
Lemma 4. Suppose that f, g : A → R are functions and α, β ∈ R. If f and g
are summable, then αf + βg is summable and
X
X
X
(αf (a) + g(a)) = α
f (a) + β
g(a).
a∈A
a∈A
a∈A
P
P
Proof. Let σ1 = a∈A f (a) and σ2 = a∈A g(a), and set h = αf + βg. For
> 0, there is some F ∈ P0 (A) such that F ⊆ F ∈ P0 (A) implies that
|Sf (F ) − σ1 | < , and there is some G ∈ P0 (A) such that G ⊆ G ∈ P0 (A)
implies that |Sg (G) − σ2 | < . Let H = F ∪ G ∈ P0 (A). If H ⊆ H ∈ P0 (A),
then, as F ⊆ H and G ⊆ H,
X
(αf (a) + βg(a)) − ασ1 − βσ2 |Sh (H) − (ασ1 + βσ2 )| = a∈H
= |αSf (H) + βSg (H) − ασ1 − βσ2 |
≤
|α||Sf (H) − σ1 | + |β||Sg (H) − σ2 |
≤
|α| + |β|;
we write ≤ rather than < in the last inequality to cover the case where α = β =
0. It follows that Sh converges to ασ1 + βσ2 .
The following lemma is simple to prove and ought to be true, but should not
to be called obvious. For example, the Ces`aro sum of the sequence 1, −1, 1, −1, . . .
is 12 , while the Ces`
aro sum of the sequence 1, −1, 0, 1, −1, 0, . . . is 31 .
Lemma 5. If f : A → R is summable, then for any set C that contains A, the
function g : C → R defined by
(
f (c) c ∈ A
g(c) =
0
otherwise
is summable, and
X
f (a) =
a∈A
X
g(c).
c∈C
P
Proof. Let σ = a∈A f (a). For > 0, there is some F ∈ P0 (A) such that
F ⊆ F ∈ P0 (A) implies that |Sf (F ) − σ| < . If F ⊆ H ∈ P0 (C), then, as
6
F ⊆ H ∩ A ∈ P0 (A),
|Sg (H) − σ| =
=
=
X
g(c) − σ c∈H
X
X
g(c) +
g(c) − σ c∈H∩A
c∈H\A
X
X
f (a) +
0 − σ a∈H∩A
c∈H\A
=
|Sf (H ∩ A) − σ|
<
.
This shows that Sg converges to σ.
The previous two lemmas are useful, and also convince us that unordered
summation works similarly to finite sums. We now establish conditions under
which a function is summable.
Lemma 6. If f : A → R is nonnegative and there is some M ∈ R such that
F ∈ P0 (A) for all Sf (F ) ≤ M P
, then f is summable. If f : A → R is nonnegative
and summable, then Sf (F ) ≤ a∈A f (a) for all F ∈ P0 (A).
Proof. Suppose there is some M ∈ R such that if F ∈ P0 (A) then Sf (F ) ≤ M .
That is, M is an upper bound for the range of Sf . Because f is nonnegative,
the net Sf is increasing. We apply Lemma 3, which tells us that Sf converges
to the supremum of its range. That Sf converges
P means that f is summable.
Suppose that f is summable, and let σ = a∈A f (a). Suppose by contradiction that there is some F0 ∈ P0 (A) such that Sf (F0 ) > σ, and let = Sf (F0 )−σ.
Then there is some F ∈ P0 (A) such that F ⊆ F ∈ P0 (A) implies that
|Sf (F ) − σ| < . As F ⊆ F0 ∪ F ∈ P0 (A), we have |Sf (F0 ∪ F ) − σ| < , and
hence
Sf (F0 ∪ F ) < σ + = Sf (F0 ).
But F0 is contained in F0 ∪ F and f is nonnegative, so
Sf (F0 ) ≤ Sf (F0 ∪ F ),
which gives Sf (F0 ) < Sf (F0 ), a contradiction. Therefore, there is no F ∈ P0 (A)
for which Sf (F0 ) > σ.
Lemma 7. Suppose that f : A → R is a function and that A+ = {a ∈ A :
f (a) ≥ 0} and A− = {a ∈ A : f (a) ≤ 0}. Then, f is summable if and only if f
is summable over both A+ and A− . If f is summable, then
X
X
X
f (a) =
f (a) +
f (a).
a∈A
a∈A+
a∈A−
7
Proof. Suppose that f is summable. Because f is summable, there is some
E ∈ P0 (A) such that E ⊆ F ∈ P0 (A) implies that |Sf (F ) − σ| < 1. Define
E+ = {a ∈ E : f (a) ≥ 0} ∈ P0 (A+ ),
E− = {a ∈ E : f (a) ≤ 0} ∈ P0 (A− ).
If G ∈ P0 (A+ ) then E ⊆ G ∪ E ∈ P0 (A), and hence |Sf (G ∪ E) − σ| < 1. We
have
X
X
X
X
Sf+ (G) =
f (a) ≤
f (a) =
f (a) −
f (a),
a∈G
a∈G∪E+
a∈G∪E
a∈E−
and hence
Sf+ (G) ≤ Sf (G ∪ E) − Sf (E− ) < σ + 1 − Sf (E− ).
That is, σ + 1 − Sf (E− ) is an upper bound for the range of Sf+ . The net
Sf+ is increasing, hence applying Lemma 3 we get that Sf+ converges. That
is, f+ is summable. If H ∈ P0 (A− ), then E ⊆ H ∪ E ∈ P0 (A), and hence
|Sf (H ∪ E) − σ| < 1. We have
X
X
X
X
Sf− (H) =
f (a) ≥
f (a) =
f (a) −
f (a),
a∈H
a∈H∪E−
a∈H∪E
a∈E+
and then
Sf− (H) ≥ Sf (H ∪ E) − Sf (E+ ) > σ − 1 − Sf (E+ ),
showing that −σ + 1 + Sf (E+ ) is an upper bound for the net −Sf− . As −Sf−
is increasing, by Lemma 3 it converges, and it follows that Sf− converges. That
is, f− is summable.
Suppose that f is summable over both A+ and A− . Let f+ be the restriction
of f to A+ and let f+ be the restriction of f to A+ , and define g+ , g− : A → R
by
(
(
f (a) a ∈ A+
0
a ∈ A+
g+ (a) =
g− (a) =
0
a ∈ A− ,
f (a) a ∈ A− .
By Lemma 5, f+ being summable implies that g+ is summable, with
X
X
f+ (a) =
g+ (a),
a∈A+
a∈A
and f− being summable implies that g− is summable, with
X
X
f− (a) =
g− (a).
a∈A−
a∈A
But f = g+ + g− , so by Lemma 4 we get that f is summable, with
X
X
X
X
X
f (a) =
g+ (a) +
g− (a) =
f+ (a) +
f− (a).
a∈A
a∈A
a∈A
a∈A+
8
a∈A−
If f : A → R is a function, we define |f | : A → R by |f |(a) = |f (a)|.
Theorem 8. If f : A → R is a function, then f is summable if and only if |f |
is summable.
Proof. Let A+ = {a ∈ A : f (a) ≥ 0} and A− = {a ∈ A : f (a) ≤ 0}, and let f+
and f− be the restrictions of f to A+ and A− respectively. Suppose that f is
summable. Then by Lemma 7 we get that f+ is summable and f− is summable.
Let F ∈ P0 (A) and write F+ = {a ∈ F : f (a) ≥ 0}, F− = {a ∈ F : f (a) ≤ 0}.
We have
X
X
X
S|f | (F ) =
|f (a)| =
f (a) −
f (a) = Sf+ (F+ ) − Sf− (F− ).
a∈F
a∈F+
a∈F−
P
But by Lemma 6, because the net Sf+ is increasing we have Sf+ (F+ ) ≤ a∈A+ f+ (a),
P
and because the net −Sf− is increasing we have −Sf− (F− ) ≤ − a∈A− f− (a).
P
P
Therefore, a∈A+ f+ (a)− a∈A− f− (a) is an upper bound for the range of S|f | .
Moreover, S|f | is increasing, so by Lemma 6 it follows that S|f | converges, i.e.
that |f | is summable.
Suppose that |f | is summable. By Lemma 6, for any F ∈ P0 (A+ ) we have
X
Sf+ (F ) = S|f | (F ) ≤
|f |(a),
a∈A
P
i.e., a∈A |f |(a) is an upper bound for the range of Sf+ . As Sf+ is increasing,
by Lemma 6 it follows that Sf+ converges, i.e., that f+ is summable. Because
−Sf− is increasing, we likewise get that −Sf− converges and hence that Sf−
converges, i.e. that f− is summable. Now applying Lemma 7, we get that f is
summable.
Theorem 9. If f : A → R is summable, then {a ∈ A : f (a) 6= 0} is countable.
Proof. Suppose by contradiction that {a ∈ A : f (a) 6= 0} is uncountable. We
have
[
1
{a ∈ A : f (a) 6= 0} = {a ∈ A : |f (a)| > 0} =
a ∈ A : |f (a)| ≥
.
n
n∈N
Since this is a countable union, there is some n ∈ N such that a ∈ A : |f (a)| ≥ n1
is uncountable; in particular, this set is infinite. Because f is summable, by Theorem 8 we have that |f | is summable, with unordered sum σ. Hence, there is
some F1 ∈ P0 (A) such that
F1 ⊆ F ∈ P0 (A)
implies that |S|f | (F ) − σ| < 1.
Let F be a finite subset of a ∈ A : |f (a)| ≥ n1 with at least n(σ + 1) elements.
Then
X
X
1
S|f | (F ∪ F1 ) =
|f (a)| ≥
|f (a)| ≥ n(σ + 1) · = σ + 1.
n
a∈F ∪F1
a∈F
But F1 ⊆ F ∪ F1 ∈ P0 (A), so S|f | (F ∪ F1 ) < σ + 1, a contradiction. Therefore,
{a ∈ A : f (a) 6= 0} is countable.
9
8
References
McArthur [52]
Schaefer [64, p. 120]
Roytvarf [63, p. 282]
McShane [53]
Diestel [13, Chapter VI]
Remmert [59, p. 29]
Sorenson [67]
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Kadets and Kadets [40]
Manning [50]
Bottazini [2]
Boyer [4]
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Smithies [66]
Dugac [16]
Grattan-Guinness [31] and [33] and [32] and [30]
Whiteside [75]
Schaefer [65]
Cauchy [5]
Polya [57]
Lakatos [44]
Krantz [43]
Cunha [12]
Youschkevitch [77]
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9
Probability
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10
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